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Proof step 4 from proof mapex-new (previous step)

P(H) ⊦ r = null, r ∈ H, ∃ o. ((r => o * P))(H ++ r)

User applied rule exists right

And obtained the following new goals

Proof step 5:
P(H) ⊦ r = null, ((r => (H ++ r)[r] * P))(H ++ r), r ∈ H

Rule argument

pos: right 3 terms: (H ++ r)[r] abort: F discard: T computed: F

Additional information

Used terms: (H ++ r)[r]